Engineering Mechanics

Chapter 5 : Equilibrium of Forces „„„„„ 75

Solution. Given : Load at D (P) = 4 kN ; Load at E (W) = 2 kN; a = 3·0 m ; b = 0·9 m
and c = 1.8 m

Now let us use the conditions of equilibrium one by one.

HC= RB ...(For ∑H = 0)

VC = P + W = 4 + 2 = 6 kN ...(For ∑V = 0)

and RB.C= P.a + W.b ...(For ∑M = 0 about C)

∴ RB × 1.8 = (4 × 3.0) + (2 × 0.9) = 13.8

or

13.8

7.67 kN

B 1.8

R == Ans.

and HC= RB = 7.67 kN Ans.

Now the reaction at C,

(^22) 7.67 (^226) 9.74 kN.
RHVCCC=+= += Ans
EXERCISE 5.2

1. A spherical ball of weight 50 N is suspended vertically by a string 500 mm long. Find the
   magnitude and direction of the least force, which can hold the ball 100 mm above the
   lowest point. Also find tension in the string at that point. [Ans. 30 N ; 40 N]
   Hint. The force will be least, when it is applied at an angle of 90° with the string.


2. A jib crane shown in Fig. 5.32 is required to lift a load of 5 kN. Find, graphically, the
   forces in the jib and tie. Also check the answer analytically.
   [Ans. 13.7 kN (tension) ; 9.7 kN (compression)]

Fig. 5.32. Fig. 5.33.

3. Two smooth spheres of weight W and radius r each are in equilibrium in a horizontal
   channel of A and B vertical sides as shown in Fig. 5.33.
   Find the force exerted by each sphere on the other. Calculate these values, if r = 250 mm,
   b = 900 mm and W = 100 N. [Ans. 133.3 N ; 166.7 N ; 133.3 N ; 200 N]